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Maximum power transfer question !

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dougy83 said:
I can't read into the future.

YOu can't find any problem with my final math either, apparently. But let's finish the math anyway. For RS=10 ohms, we have P=V^2/40, as previously stated. Now, change RS= 1 ohm, then the power goes up to V^2/12 for RL=10 ohms. BUT, if we match RL to RS, then the power is P=V^2, which is again, a maximum.
 
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dougy83,

I think both of us are embarrassed by some individuals in this group because we have to devote so much bandwidth to explain and convince some folks, including one in particular, of a really simple principle. Even when proven by simple mathematics, it is not convincing to some. As you have found out, it is not worth devoting attention to the willfully ignorant. It is encouraging to note that at least you have the perspicacity to understand and agree to what I am saying.

Ratch
 
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Simple math proves what most members already knew, that minimising RS does not make for maximum power transfer (post #42). But some intentionally, admittedly wont' read the proof. So now who's willfully ignorant?
 
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hi,
I think where people are getting hung upon is the phrase maximum power transfer.

Maximum power transfer means transferring the maximum available power from the source to the load.

eg:
If a Voltage source of say 10V with an internal resistance of say 10R, this means the maximum available power is 10Watts.

So to transfer the maximum amount of power to the load, the load resistance has to be 10R.

So when the load resistance is equal to the source resistance, in this case, the total loop resistance is 20R.

This means that 0.5A flows [exists] in the loop, so the power dissipated in the source and load resistance equals 2.5Watts each.
Thats a total of 5Watts from a possible 10W, so the transfer is only 50% efficient.

If the Load resistance is lower than the source resistance more power power will be dissipated by the source resistance
If the Load resistance is higher than the source resistance less current will flow in the loop so the power dissipated by the the load will be lower

The above DC, resistive, explanation should answer the OP question

To maximise the power dissipation in a load, the source resistance should be lower than the load resistance.

If you dont agree, consider the low output impedance of an audio amp, say 0.02R driving a 4R speaker.
 
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Hi Eric,

I agree with everything you wrote. But, I believe the OP wants to know about maximum power transfer, which he titled the thread. This important concept sets a limit on the maximum power that can be delivered to a load at 1/2 the available power. For example, if an output amp of 0.2R is driving a 4R speaker, and the peak output voltage is, say 10V, then the power delivered to the load is 10^2/4.2^2*4 = 22.7W. But the maximum available power ( realized when RL=RS ) is 10^2/.4R=250W. So, this system only delivers 22.7/250=9% of the available power. The maximum power possible that can be delivered to a load would be 1/2 the max power available, and that's the important point the OP is asking about. Of course that isn't practical in a amp/speaker, and some sharp members already pointed that out. But there are scenarios where the maximum power rule is practical and important.

The other important point is that one member proved the above with simple, elegant math. I defend his proof because it was correct and crucial to understanding the concept. Had the thread been able to actually stay on topic, the confusion wouldn't happen.
 
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hi D,
I did cover the OP's question about maximum power transfer.:)

I didn't pay too much attention to all the maths which this thread spawned, its not rocket science.

The calculations are basic derivations of Ohm and Power Laws.

The two basic points are:
That to get maximum power transfer from the Source to a Load their resistances must be the same.

And to get maximum power into a Load ideally the Source resistance should be zero.

BTW: I know that you agree.:D
Eric
 
ericgibbs,

And to get maximum power into a Load ideally the Source resistance should be zero.

Correct! That is what I have been saying all along. For a fixed load, the closer the source resistance is to zero, the more power is transferred. I proved it mathematically and by example, and my Wikileaks post also agrees. Certain detractors keep trying to say otherwise by changing the load impedance. It does not matter what the OP asked about, or whether it is practical to change the source impedance. The fact remains that for a fixed load, one can transfer more power if the source impedance is the lowest possible value instead of being equal to the load impedance. I cannot figure out why some folks have such a hard time understanding that. Like you said, it is not rocket science.

Ratch
 
Nobody has a hard time understanding the trival case of decreasing the source resistance. But it does matter what the OP asked about. That's the purpose of the forums. Further, decreasing the source resistance only does not produce maximum power transfer, as proved by several members in several ways.
 
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I cannot figure out why some folks have such a hard time understanding that. Like you said, it is not rocket science.

As has been pointed out to you endlessly, you're not answering the original question - you're answering something completely different - why can't you understand that?.
 
Nigel Goodwin,

As has been pointed out to you endlessly, you're not answering the original question - you're answering something completely different - why can't you understand that?.

The OP said:
while studying i had this odd question. we know that the maximum power transfer theorm indicates that impedence matching is necessery to achieve max power transfer to the load, but when it comes to amplifier networks, and studying the effect of load resistance, we say that the load resistance better be large to achieve higher gain, so how do these two facts get along ?

i'm saying that creating higher load for amplifier networks establishes some change in current gain to keep power to its max maybe ?

I pointed out something very, very closely related to what the OP asked. Specifically that more power could be transferred if the source resistance were lowered to as low a value as possible. What I said was not off into la-la land. It was pertinent and on the mark. It was also correct. Nevertheless, many folks could not process that, and say something like "That's right, I never thought of that." . Instead, they said it was wrong, which is untrue. So to answer your question, no, I did not answer something "different". The OP asked about the ratios of the source and load impedances, so my post was on the mark.

Ratch
 
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